Would someone please check my answers? thanks
During a lie detector test, a voltage of 6 V is impressed across two fingers. When a certain question is asked, the resistance between the fingers drops from 400,000 ohms to 200,000 ohms. What is the current (a)initially through the fingers, and (b) when the resistance between them drops?
(a).000015 amps
(b).00003 amps
Using I = V/R, I agree with your answers.
To find the current initially and when the resistance drops, we'll use Ohm's Law, which states that the current flowing through a conductor is directly proportional to the voltage impressed across it and inversely proportional to its resistance.
(a) To find the initial current flowing through the fingers, we can use Ohm's Law:
I = V / R
Where:
I = Current (in amperes)
V = Voltage (in volts)
R = Resistance (in ohms)
Given:
V = 6 V
R = 400,000 ohms
Substituting the values into the equation:
I = 6 V / 400,000 ohms
I = 0.000015 amps
Hence, the initial current through the fingers is 0.000015 amps.
(b) Now let's calculate the current when the resistance between the fingers drops:
Given:
V = 6 V
R = 200,000 ohms
Using Ohm's Law:
I = V / R
Substituting the values:
I = 6 V / 200,000 ohms
I = 0.00003 amps
Therefore, when the resistance between the fingers drops, the current through them becomes 0.00003 amps.
So, the correct answers are:
(a) Initially, the current through the fingers is 0.000015 amps.
(b) When the resistance drops, the current through the fingers is 0.00003 amps.